Bill kicks a rock off the top of his apartment
building. It strikes a window across the street
12 m away.
The acceleration of gravity is 9.8 m/s2 .
If the window is 24 m below the position
where Bill contacted the rock, how long was
it in the air?
Answer in units of s.
d =0.5gt^2.
24 = 0.05 * 9.8t^2,
24 = 4.9t^2,
t^2 = 24 / 4.9 = 4.9,
t = sqrt(4.9) = 2.2 s.
To find the time the rock was in the air, we can use the equation of motion for vertical motion:
s = ut + (1/2)at^2
Where:
- s is the displacement (distance) traveled by the object
- u is the initial velocity of the object
- a is the acceleration
- t is the time
In this case, we are looking for the time (t). The initial velocity (u) can be assumed to be zero since Bill only kicks the rock off the building, meaning it starts at rest. The acceleration (a) is the acceleration due to gravity (9.8 m/s^2).
Now, let's analyze the given information. The rock strikes a window 12 m away horizontally and the window is 24 m below the position where Bill contacted the rock. Since the horizontal distance does not affect the time of flight, we only need to focus on the vertical motion.
Using the equation s = ut + (1/2)at^2 and substituting the known values:
- s = -24 m (negative sign indicates downward direction)
- u = 0 m/s
- a = 9.8 m/s^2
We can solve for t. Rearranging the equation, we get:
-24 = (1/2)(9.8)t^2
Multiplying both sides by 2:
-48 = 9.8t^2
Dividing both sides by 9.8:
t^2 = -48 / 9.8
Taking the square root of both sides to solve for t:
t = √(-48 / 9.8)
Since time cannot be negative in this context, we can ignore the negative square root. Thus:
t ≈ 2.020 seconds
Hence, the rock was in the air for approximately 2.020 seconds.